Willis Tran
07/02/2024 · High School
Find the derivative of: \( -4 \sin ^{2}\left(-5 x^{5}\right) \). Hint: \( \sin ^{2}(x)=[\sin (x)]^{2} \ldots \) so use the chain rule (twice!).
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To find the derivative of the function \( y = -4 \sin^{2}(-5x^{5}) \), we will use the chain rule twice.
First, let's rewrite the function for clarity:
\[
y = -4 [\sin(-5x^{5})]^{2}
\]
Now, we will differentiate \( y \) with respect to \( x \).
1. **Differentiate the outer function**: The outer function is \( -4u^2 \) where \( u = \sin(-5x^{5}) \). The derivative of \( -4u^2 \) with respect to \( u \) is:
\[
\frac{dy}{du} = -8u
\]
2. **Differentiate the inner function**: The inner function is \( u = \sin(-5x^{5}) \). The derivative of \( \sin(v) \) is \( \cos(v) \cdot \frac{dv}{dx} \), where \( v = -5x^{5} \). Thus, we need to find \( \frac{du}{dx} \):
\[
\frac{du}{dx} = \cos(-5x^{5}) \cdot \frac{d}{dx}(-5x^{5})
\]
Now, we compute \( \frac{d}{dx}(-5x^{5}) \):
\[
\frac{d}{dx}(-5x^{5}) = -25x^{4}
\]
So, we have:
\[
\frac{du}{dx} = \cos(-5x^{5}) \cdot (-25x^{4}) = -25x^{4} \cos(-5x^{5})
\]
3. **Combine using the chain rule**: Now we can apply the chain rule:
\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = (-8u) \cdot (-25x^{4} \cos(-5x^{5}))
\]
Substituting back \( u = \sin(-5x^{5}) \):
\[
\frac{dy}{dx} = -8 \sin(-5x^{5}) \cdot (-25x^{4} \cos(-5x^{5}))
\]
This simplifies to:
\[
\frac{dy}{dx} = 200 x^{4} \sin(-5x^{5}) \cos(-5x^{5})
\]
Thus, the derivative of \( -4 \sin^{2}(-5x^{5}) \) is:
\[
\boxed{200 x^{4} \sin(-5x^{5}) \cos(-5x^{5})}
\]
Quick Answer
The derivative of \( -4 \sin^{2}(-5x^{5}) \) is \( 200 x^{4} \sin(-5x^{5}) \cos(-5x^{5}) \).
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